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Assessing Renal Function: Renal Measurements

MEASURING BLOOD VOLUME
6 - 8 % OF BODY WEIGHT IN Kg

Dye Dilution Method for Calculating Fluid Volumes:

PLASMA VOLUME

VOLUME = __AMOUNT mg_____
CONCENTRATION mg / ml

USE A SUBSTANCE: > 69,000 MW
NOT METABOLIZED

BLOOD VOLUME: CORRECT PLASMA VOLUME (PV) FOR HEMATOCRIT (HCT)

BLOOD VOL = P V x 1
1 - HCT

INTERSTITIAL FLUID VOLUME

VOL = AMT / CONC
USE INULIN
• 5,200 MW
• NOT METABOLIZED
• NOT TAKEN UP BY CELLS

TOTAL BODY WATER

USE D₂O or ³H₂O
ALLOW TIME FOR DISTRIBUTION AMONG COMPARTMENTS
USE USUAL EQUATION:
VOL = AMT / CONC

Renal Function: Glomerular Filtration, Renal Clearance, & Renal Blood Flow

Calculating Renal Blood Flow and GFR

To calculate Clearance & RBF,
• Inject known amount of a solute, i.v.
• Wait a few minutes
• Measure P_A in any artery
• Measure P_V in RENAL VEIN
• Measure Rate of urine formation
• Measure [solute]_urine

Renal clearance:

volume of plasma per minute needed to excrete the quantity of solute appearing in the urine in a minute

If there were 1 mg of solute Z in 100 ml of plasma, and you found 0.5 mg of Z appearing in the urine/ min, then the clearance of Z would = 50 ml of plasma

Renal Clearance:
The hypothetical volume of plasma from which a substance is completely removed per minute in one pass thorough the kidney

Figure 17.19 may help clarify or visualize these processes!

Clearance
C = U x V / P_A

• U = [solute] in urine (mg/ml)
• V = volume of urine/min (ml/min)
• P_A = [solute] in arterial plasma (mg/100 ml plasma

To Calculate Clearance
P_A = 1.0 mg/100 ml plasma
U = 0.1 mg/ml
V = 1.0 ml/min
C = 0.1mg/ml x 1.0 ml/min
mg / 100 ml
C = 10 ml/min

Understanding Clearance:
If in this example, 0.1 mg of solute appears in urine / min, in how much plasma was that 0.1 mg delivered if PA = 1.0 mg/100 ml ?
Answer = 10 ml/min

In other words, that 0.1 mg of solute that appeared in the urine/min was dissolved in 10 ml of plasma. Thus C = 10 mls/min.

Extraction Ratio (E)

The E of a solute is equal to the fraction of the substance that is removed from the plasma in one pass through the kidney:
E = P_A - P_V / P_A

Calculate E
P_A = 0.2 mg/100 ml
P_V = 0.1 mg/100 ml
E = 0.2 - 0.1 / 0.2
E = 0.5

Renal Plasma Flow
RPF = C / E
RPF = 10 ml/min / 0.5
RPF = 20 ml / min
Renal Blood Flow
RBF = RPF x 1 / 1- hct
If hct = 0.5,
RBF = 20 ml/min x 1 / 1 - 0.5
RBF = 40 ml / min

Calculating Renal BF:
Substance P_A P_V U V

X 1.0 mg/100ml 0.8/100 0.2 mg/ml 1 ml/min

Y 1.0 0.3 0.7 1

Z 1.0 0.9 0.1 1

Hct = 0.50

To calculate RBF, first calculate C and E of substance with largest E
E_Y = 0.7
C_Y = 0.7 mg/ml x 1 ml/min
1 mg / 100 mls

C_Y = 70 ml / min

From C and E, calculate RPF:
RPF = C / E
RPF = 70 ml / min
0.7
RPF = 100 ml/ min

From RPF & Hct, calculate RBF:
RBF = RPF x 1 / 1 - Hct
RBF = 100 ml / min x 1 / 0.5

RBF = 200 ml / min

GFR = C of solute which is ONLY FILTERED, e.g., inulin.
WHY? See Figure 17.20 to see this concept.

Inferences from GFR
• If C_solute = GFR, solute is only filtered
• If C_solute > GFR, solute is filtered and secreted
• If C_solute < GFR, solute is filtered and reabsorbed

Figures 17.20 & 21 (a & b) are excellent representations of these concepts !

Filtration Fraction
FF = fraction of plasma that is made into filtrate
Usually = 0.2 in humans

Calculate:
FF = GFR / RPF

Inferences from FF:
• If E_solute = FF, solute is only filtered
• If E_solute > FF, solute is filtered and secreted
• If E_solute < FF, solute is filtered and reabsorbed

Assume X from data set (above) is only filtered

Then, GFR = C_X = 0.2 x 1
1/ 100
GFR = 20 ml / min

FF = GFR / RPF
FF = 20 ml/min / 100 ml/min
FF = .2

Calculate Amount of a solute filtered / min:
Amount Filtered / min = P_A x GFR
For Z, Amt Filt =
1.0mg/100 mls x 20 ml / min
Amt Filt = 0.2 mg / min

Amount of a Solute Appearing in Urine / min
Amt Urine = U x V
For Solute Z, Amt Urine = 1 mg / ml x 1 ml / min
Amt in Urine / min = 0.1 mg / min

Amount reabsorbed / min
= Amt filtered - Amt in Urine
= (GFR x PA) - (U x V)
= (20 ml/min x 1.0 / 100) – 1 mg/ml x 1 ml/min)
= (0.2 mg/min) - (0.1 mg / min)
= 0.1 mg / min is reabsorbed

Amount Secreted / min
= Amt in Urine - Amt Filtered
= (U x V) - (PA x GFR)
= (0.7 mg/ml x 1 ml/min) – (1 mg/ 100 ml x 20 ml/min)
= (0.7 mg/min) - (0.2 mg/min)
= 0.5 mg/min

Inferences:
If UV = GFR P_A , solute is only filtered
If UV > GFR P_A , solute is filtered and secreted
If UV < GFR P_A , solute is filtered and reabsorbed